Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 273

\(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}=\mathrm{C}^{-}\), Then \(\mathrm{C}^{\rightarrow}\) is perpendicular to (A) \(\mathrm{A}^{\rightarrow}\) only (B) \(\mathrm{B}^{\rightarrow}\) only (C) \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) both when the angle between them is \(\ldots\) (D) \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) both whatever to be the angle between them

Short Answer

Expert verified
The correct answer is (D): \(C^{\rightarrow}\) is perpendicular to both \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\), irrespective of the angle between them, because it is a result of their cross product.

Step by step solution

01

Recognize the properties of the cross product

The cross product of two vectors A and B, denoted as \(A \times B\), results in a new vector C, which has the following properties: 1. The magnitude of C is equal to the product of the magnitudes of A and B and the sine of the angle between them: \(|C| = |A||B|\sin(\theta)\). 2. Vector C is perpendicular to both A and B. So the statement states that the cross product of A and B results in vector C, and C is negative.
02

Analyze the options

Here are the options presented in the question: (A) C is perpendicular to A only. (B) C is perpendicular to B only. (C) C is perpendicular to both A and B when the angle between them is ... (D) C is perpendicular to A and B whatever the angle between them. Since the cross product of two vectors is a vector that is perpendicular to both A and B, the only option that matches the properties of the cross product is option (D).
03

Choose the correct answer and explain

The correct answer is (D): \(C^{\rightarrow}\) is perpendicular to both \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\), irrespective of the angle between them, because it is a result of their cross product. The cross product of any two vectors always produces a vector that is perpendicular to the original vectors, regardless of the angle between the original vectors. Note that the statement does not impact the final answer since only the direction is negative, not the property of being perpendicular.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Angle of projection of a projectile with horizontal line is \(\theta\) at time \(t=0\), After what time the angle will be again \(\theta\) ? (A) \([(\mathrm{V} \cos \theta) / \mathrm{g}]\) (B) \([(\mathrm{V} \sin \theta) / \mathrm{g}]\) (C) \(\left[\left(\mathrm{V}_{0} \sin \theta\right) / 2 \mathrm{~g}\right]\) (D) \(\left[\left(2 \mathrm{~V}_{0} \sin \theta\right) / \mathrm{g}\right]\)

The relation between velocity and position of a particle is \(\mathrm{V}=\mathrm{Ax}+\mathrm{B}\) where \(\mathrm{A}\) and \(\mathrm{B}\) are constants. Acceleration of the particle is \(10 \mathrm{~ms}^{-2}\) when its velocity is \(\mathrm{V}\), How much is the acceleration when its velocity is $2 \mathrm{~V}$ (A) \(20 \mathrm{~ms}^{-2}\) (B) \(10 \mathrm{~ms}^{-1}\) (C) \(5 \mathrm{~ms}^{-2}\) (D) 0

The area under acceleration versus time graph for any time interval represents... (A) Initial velocity (B) final velocity (C) change in velocity in the time interval (D) Distance covered by the particle

The ratio of pathlength and the respective time interval is (A) Mean Velocity (B) Mean speed (C) instantaneous velocity (D) instantaneous speed

The motion of a particle along a straight line is described by the function \(\mathrm{x}=(3 \mathrm{t}-2)^{2}\). Calculate the acceleration after $10 \mathrm{~s}$. (A) \(9 \mathrm{~ms}^{-2}\) (B) \(18 \mathrm{~ms}^{-2}\) (C) \(36 \mathrm{~ms}^{-2}\) (D) \(6 \mathrm{~ms}^{-2}\)
See all solutions

Recommended explanations on English Textbooks

Cues and Conventions

Read Explanation

Textual Analysis

Read Explanation

Lexis and Semantics

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation

Prosody

Read Explanation

Creative Story

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free